Problem: Rewrite the function by completing the square. $f(x)=x^{2}-12x+50$ $f(x)=(x+$
We want to complete $x^2{-12}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-12}}{2}\right)^2={36}$ to it: $x^2{-12}x+{36}=(x-6)^2$ In order to keep the expression equivalent, we add and subtract ${36}$, not forgetting the expression's constant term, $50$ : $\begin{aligned} f(x)&=x^2-12x+50 \\\\ &=x^2-12x+{36}+50-{36} \\\\ &=(x-6)^2+50-36 \\\\ &=(x-6)^2+14 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 6)^2 + 14$ This is equivalent to $f(x)=(x+{-6})^2+14$